Question

Three numbers are chosen at random without replacement from $\{1,2,3,.....,8\}$. Probability that their minimum is equal to order of differential equation $x^3\frac{d^{3}y}{d{x}^3}+x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0$, given that their maximum is equal to degree of differential equation $y^2=\frac{dy}{dx}(a+\sqrt[2]{2{\left(\frac{dy}{dx}\right)}^3}$

• A. $\frac{3}{8}$
• B. $\frac{1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{2}{5}$

Answer 1

The correct option is B $\frac{1}{5}$

Order of the differential equation $x^3\frac{d^{3}y}{d{x}^3}+x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0$ is 3 since the maximum differential element used is 3.

Also, the degree of the equation $y^2=2\frac{dy}{dx}(x+5\sqrt{(\frac{dy}{dx}{)}^3})$ is found out by squaring both sides since the power of $\frac{dy}{dx}$ has to be an integer.

It is found out to be 6.

So, our sample space becomes to find 2 numbers from 1 to 5 and the question becomes to find out 1 number from 4 to 5.

$\Rightarrow$ probability = $\frac{{}^2{C}_1}{{}^5{C}_2}=\frac{1}{5}$